Need help in the following question: The area of a rectangle is 30 square feet. The length of the rectangle is 4 feet shorter than double the width. Find the length and the width.

Question

anonymous · Answer

Area of rectangle  = (length) x (width)

= ( (2 x width) - 4) x (width)

solve for width then plug back in to solve for length

anonymous · Answer

im not understanding, sorry.

calculusxy · Answer

Lets put this into an equation, so that we can find the length first:
w^2-4=length

anonymous · Answer

okay sounds good !

calculusxy · Answer

Now we can use guess and check to replace the 'w' variable.

anonymous · Answer

plug in any number that will equal the are of the rectangle -30 square feet?

anonymous · Answer

*area

calculusxy · Answer

Actually that will equal to the length. Becuase after the length, we can easily find the width

anonymous · Answer

oh so how can i do that ?

calculusxy · Answer

Wait, I am trying to figure it out.

anonymous · Answer

thank you

calculusxy · Answer

http://www.ehow.com/how_8472576_length-width-rectangle-given-area.html

anonymous · Answer

okay thank you i'll check this and figure what info. i have and don't have

calculusxy · Answer

No problem :)

anonymous · Answer

Use what you learned from calculusxy that the length is = 2 times the width - 4 so in equation form it would look like this with "w" representing the "width"$$length = (2w-4)$$

my earlier equation represented the definition of the area of rectangle$$Area = (length)(width)$$ plug in your length which was represented up there to get this equation$$30 = (2w-4)(w)$$ simplify it to this$$30 = 2w^2-4w$$ then to this$$30 = 2(w^2-2w)$$complete the square for w^2-2w to get:$$30+2 = 2(w^2-2w+1)$$simplify it to this$$32 = 2(w-1)^2$$solving for "w"$$16 = (w-1)^2$$$$4 = w-1$$$$w =3$$ now plug that into your original equation$$30 = (length)(3)$$ then solve for length$$length = 10$$